Lebesgue measure on $\mathbb{R}$ Let $l$ be the Lebesgue measure on $ \sigma$-algebra $L$ of Lebesgue measurable subsets of $\mathbb{R}$ and $A\in L$ with $l(A)>0$.

How do I prove that for a $0<\epsilon<1$ there exists an open $U$ such that
  $$ l(A\cap U)>\epsilon\cdot l(U) ?$$

What I have tried:
I know that the Lebesgue measure is given by
$$l(X)=\sup\{l(C):C\subset X,C\text{ compact}\}$$
or
$$l(X)=\inf\{l(O):X\subset O,O\text{ open}\}$$
Thus I can find an open $O$ with $A\subset O$ and $l(A)>l(O)-\delta$ for any $\delta$.
Taking $\delta=l(A)(1/\epsilon-1)$, we then get $l(A)>\epsilon\cdot l(O)$.
How do I continue from here?
 A: The idea to use the outer regularity (i.e. the outer approximation with open sets) of the Lebesgue measure is correct.
We are going to show that, indeed, the claim is true with $U$ open interval.
Clearly it is not restrictive to assume $l(A)$ is finite, since otherwise it is enough to consider $A \cap [k, k+1)$ for some $k\in\mathbb{Z}$.
Assume by contradiction that $l(A\cap I) \leq \epsilon l(I)$ for every open interval $I$.
Given any $\delta > 0$, by the outer approximation property we can find a sequence of pairwise disjoint open intervals $(I_k)$ such that
$$
A \subset \bigcup_k I_k,
\qquad
\sum_k l(I_k) \leq l(A) + \delta.
$$
Hence
$$
l(A) = l(A \cap \bigcup_k I_k) \leq \sum_k l(A\cap I_k)
\leq \sum_k \epsilon l(I_k) \leq \epsilon(l(A) + \delta).
$$
If $\delta$ is choosen at the beginning such that
$$
\epsilon\delta < (1-\epsilon) l(A)
\quad \text{i.e.} \quad
\delta < \frac{1-\epsilon}{\epsilon}\, l(A), 
$$
we get
$$
l(A) \leq \epsilon(l(A) + \delta)
< \epsilon l(A) + (1-\epsilon) l(A) = l(A),  
$$
a contradiction.
